Problem: $ \left(\dfrac{256}{81}\right)^{-\frac{3}{4}}$
Answer: $= \left(\dfrac{81}{256}\right)^{\frac{3}{4}}$ $= \left(\left(\dfrac{81}{256}\right)^{\frac{1}{4}}\right)^{3}$ To simplify $\left(\dfrac{81}{256}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left(? \right)^{4}=\dfrac{81}{256}$ To simplify $\left(\dfrac{81}{256}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left({\dfrac{3}{4}}\right)^{4}=\dfrac{81}{256}$ so $ \left(\dfrac{81}{256}\right)^{\frac{1}{4}}=\dfrac{3}{4}$ So $\left(\dfrac{81}{256}\right)^{\frac{3}{4}}=\left(\left(\dfrac{81}{256}\right)^{\frac{1}{4}}\right)^{3}=\left(\dfrac{3}{4}\right)^{3}$ $= \left(\dfrac{3}{4}\right)\cdot\left(\dfrac{3}{4}\right)\cdot \left(\dfrac{3}{4}\right)$ $= \dfrac{9}{16}\cdot\left(\dfrac{3}{4}\right)$ $= \dfrac{27}{64}$